Commutative Algebra/Basics on prime and maximal ideals and local rings

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Prime ideals[edit]

Definition 12.1:

Let be a ring. A prime ideal is an ideal of such that whenever , either or .

Lemma 12.2:

Let be a ring and an ideal. is prime if and only if is an integral domain.

Proof:

prime is equivalent to . This is equivalent to

.

Theorem 12.3:

Let be multiplicatively closed. Then there exists a prime ideal not intersecting .

Proof:

Order all ideals of not intersecting by set inclusion, and let a chain

be given. The ideal

(this is an ideal, since , hence , ) is an upper bound of the chain, since cannot intersect for else one of the would intersect . Since the given chain was arbitrary, Zorn's lemma implies the existence of a maximal ideal among all ideals not intersecting . This ideal shall be called ; we prove that it is prime.

Let , and assume for contradiction that and . Then , are strict superideals of and hence intersect , that is,

,
,

, , . Then , contradiction.

Projection to the quotient ring[edit]

In this section, we want to fix a notiation. Let be a ring and an ideal. Then we may form the quotient ring consisting of the elements of the form , . Throughout the book, we shall use the following notation for the canonical projection :

Definition 12.4:

Let an ideal. The map

is the canonical projection of to .

Maximal ideals[edit]

Definition 12.5:

Let be a ring. A maximal ideal of is an ideal that is not the whole ring, and there is no proper ideal such that .

Lemma 12.6:

An ideal is maximal iff is a field.

Proof:

A ring is a field if and only if its only proper ideal is the zero ideal. For, in a field, every nonzero ideal contains , and if is not a field, it contains a non-unit , and then does not contain .

By the correspondence given by the correspondence theorem, corresponds to , the zero ideal of corresponds to , and any ideal strictly in between corresponds to an ideal such that . Hence, is a field if and only if there are no proper ideals strictly containing .

Lemma 12.7:

Any maximal ideal is prime.

Proof 1:

If is a ring, maximal, then is a field. Hence is an integral domain, hence is prime.

Proof 2:

Let be maximal. Let . Assume . Then for suitable , . But then .

Theorem 12.8:

Let be a ring and an ideal not equal to all of . Then there exists a maximal with .

Proof:

We order the set of all ideals such that and by inclusion. Let

be a chain of those ideals. Then set

.

Clearly, all are contained within . Since , . Further, assume . Then for some , contradiction. Hence, is a proper ideal such that , and hence an upper bound for the given chain. Since the given chain was arbitrary, we may apply Zorn's lemma to obtain the existence of a maximal element with respect to inclusion. This ideal must then be maximal, for any proper superideal also contains .

Lemma 12.9:

Let be a ring, . Then via , maximal ideals of containing correspond to maximal ideals of .

Proof: From the correspondence theorem.

Local rings[edit]

Definition 12.10:

A local ring is a ring that has exactly one maximal ideal.

Theorem 12.11 (characterisation of local rings):

Let be a ring. The following are equivalent:

  1. is a local ring.
  2. If is a unit, then either or is a unit, where arbitrary.
  3. The set of all non-units forms a maximal ideal.
  4. If where is a unit, then one of the is a unit.
  5. If is arbitrary, either or is a unit.

Proof:

1. 2.: Assume and are both non-units. Then and are proper ideals of and hence they are contained in some maximal ideal of by theorem 12.7. But there is only one maximal ideal of , and hence , thus . Maximal ideals can not contain units.

2. 3.: The sum of two non-units is a non-unit, and if is a non-unit and , is a non-unit (for if , is an inverse of ). Hence, all non-units form an ideal. Any proper ideal of contains only non-units, hence this ideal is maximal.

3. 4.: Assume the are all non-units. Since the non-units form an ideal, is contained in that ideal of non-units, contradiction.

4. 5.: Assume , are non-units. Then is a non-unit, contradiction.

5. 1.: Let two distinct maximal ideals. Then , hence , , , that is, . is not a unit, so is, contradiction.

Localisation at prime ideals[edit]

In chapter 9, we had seen how to localise a ring at a multiplicatively closed subset . An important special case is , where is a prime ideal.

Lemma 12.12:

Let be a prime ideal of a ring. Then is multiplicatively closed.

Proof: Let . Then can't be in , hence .

Definition 12.13:

Let be a prime ideal of a ring. Set . Then

is called the localisation of at .

Theorem 12.14:

Let be a ring, be prime. is a local ring.

Proof:

Set , then . Set

.

All elements of are non-units, and all elements of are of the form , , and thus are units. Further, is an ideal since is and by definition of addition and multiplication in and since is multiplicatively closed. Hence is a local ring.

This finally explains why we speak of localisation.